Question

The 'Spin only' Magnetic moment for \([Ni(NH_3​)_6​]^{2+}\) is _____ × \(10^{−1}\) BM. (given = Atomic number of Ni : 28)

Answer 1

Correct Answer: 28

Problem Statement:
Ammonia (NH₃), functioning as a Weak Field Ligand (WFL) with Ni²⁺, results in the complex \([Ni(NH₃)_6]^{2+}\) exhibiting \( sp^3d^2 \) hybridization.

Electronic Configuration of Ni²⁺:
The electron configuration for \( Ni^{2+} \) (atomic number 28) is \( Ni^{2+} = 3d^8 \), indicating an empty 4s orbital and eight electrons in the 3d orbital.

Electron Configuration Diagram:
The 3d orbitals of \( Ni^{2+} \) are filled as follows, representing the electron arrangement:

\[ \text{Ni}^{2+} = \text{3d}^8: \quad \uparrow\downarrow \ \uparrow\downarrow \ \uparrow\downarrow \ \uparrow \ \uparrow \]

Hybridization:
The complex \([Ni(NH₃)_6]^{2+}\) displays \( sp^3d^2 \) hybridization, attributable to the presence of six NH₃ ligands requiring six available orbitals (one each from 3d, 4s, and 4p) for bond formation.

Number of Unpaired Electrons:
Two unpaired electrons reside in the 3d orbitals, influencing the complex's magnetic properties.

Magnetic Moment Calculation:
The magnetic moment \( \mu \) is calculated using the formula \( \mu = \sqrt{n(n + 2)} \, \text{BM} \), where \( n \) represents the number of unpaired electrons. With \( n = 2 \), the calculation is \( \mu = \sqrt{2(2 + 2)} = \sqrt{8} = 2.82 \, \text{BM} \). Therefore, the magnetic moment is \( 2.82 \, \text{BM} \).

Conclusion:
The calculated magnetic moment is \( \mu = 28.2 \times 10^{-1} \, \text{BM} \), corresponding to \( x = 28 \).